(Fundamental) Physics of Elementary Particlesphysics1.howard.edu/~thubsch/FPP1/Slides/1031.pdf · Department of Physics and Astronomy Howard University, Washington DC ... Fundamental

(Fundamental) Physicsof Elementary Particles

Non-abelian gauge symmetry; QCD Lagrangian, color-conservation law and equations of motion

Tristan HübschDepartment of Physics and Astronomy

Howard University, Washington DCPrirodno-Matematički Fakultet

Univerzitet u Novom Sadu

Tuesday, November 1, 11

Program

Fundamental Physics of Elementary Particles

2

A gauge (local) symmetry principle recap!e partial derivative vs. the gauge-covariant derivativeGeneral transformation “rules”

!e SU(3)c transformationsColor as a 3-dimensional chargeMatrix-valued phases and local symmetryMatrix representations of SU(3)

!e SU(3)c-invariant Lagrangian!e curvature tensor and the Bianchi identityEquations of motionColor conservation and equation of continuity


external

internal

Partial vs. gauge-covariant derivatives

Gauge (Local) Symmetry Principle

First and foremost:!e mathematical object, Ψ(r, t), used to represent a particledepends on (is a function of)

the position in space and time,the phase (as a complex function),additional degrees of freedom

spinisospincolor…

Gauge (local symmetry) principle:internal coordinates are free to depend on external ones.

3

…all of which are “coordinates,” ina broad sense ofthe word.}

base

!ber

!ber bundles/sheaves/…


connexion 1-form


Derivatives compute the rate of changeSo, if Ψ(r, t) also depends on a (r, t)-dependent phase,then the rate of change stems from

varying Ψ(r, t) explicitly, andvarying Ψ(r, t) implicitly, via varying its phase.

If “gauge” refers to “#xing” that phase,…then a “gauge-covariant” derivative…must contain two terms: Dμ = ∂μ + Aμ(r, t) ,where Aμ(r, t) is the gauge potential, a.k.a. connexion.Mathematicians: dxμDμ = dxμ∂μ + dxμAμ(r, t)

is now the (external) coordinate-independent de!nition.

4

Partial vs. gauge-covariant derivatives


General transformation “rules”


!e unitary gauge (local symmetry) transformation is of the form Uφ = exp{i φ·Q}, (Uφ† = Uφ–1)

where φ is the (array of) gauge parameter(s),where Q is the (array of) gauge transformation generator(s).

!en,Ψ(r, t) → Uφ Ψ(r, t);Ψ†(r, t) → Ψ†(r, t) Uφ–1;O(r, t) → Uφ O(r, t) Uφ–1;…and therefore also: Dμ → Uφ Dμ Uφ–1.

us, ∂μ + Aμ(r, t) → Uφ (∂μ + Aμ(r, t)) Uφ–1 implies that Aμ(r, t) → Uφ (–i (∂μφ) + Aμ(r, t)) Uφ–1 .

5


Color as a 3-dimensional charge

The SU(3)c TransformationsRecall:

Δ++ = (uuu),Δ– = (ddd),Ω– = (sss).

It follows that:either quarks are not fermions (O.W. Greenberg, 1964),

…or…6

} Spin-³/₂ baryonsS-states; no orbital angular momentumSpatially symmetric wave-functions

⇥bi , b †

j⇤= dij,

⇥bi , b j

⇤= 0 = [b †

i , b †

j ], bosons,�†

� † †

�˜fi,a ,

˜f †

j,a = dij,

�˜fi,a ,

˜f j,a = 0 =

�˜f †

i,a ,

˜f †

j,a

,⇥˜fi,a ,

˜f †

j,b⇤= dij,

⇥˜fi,a ,

˜f j,b⇤= 0 =

⇥˜f †

i,a ,

˜f †

j,a⇤, a 6= b,

)para-fermions,

⇥ ⇤ ⇥ ⇤�fi , f †

j = dij,

�fi , f j

= 0 =

�f †

i , f †

j

, fermions,


The SU(3)c TransformationsQuarks are fermions,…but have an additional degree of freedom.

January 1965: Boris V. Struminsky, Dubna (Moscow, Russia)…then with N. Bogolyubov + Albert TavchelidzeMay 1965, A. Tavchelidze: ICTP, Trieste (Italy)December 1965, Moo-Young Han + Yoichiro Nambu

integrally charged, colored quarks + 8 (color-anticolor) gluonsFinal version (w/fractionally charged quarks):1974, William Bardeen, Harald Fritzsch & Murray Gell-MannQuark: Ψn

αA(r, t), where:n = u, d, s, c, … indicates the $avorα = red, blue, yellow indicates the “color”A = 1, 2, 3, 4 indicates the component of the Dirac spinor

P.S.: Greenberg subsequently proved equivalence…7

Color as a 3-dimensional charge

meson baryon


Matrix-valued phases and local symmetry

The SU(3)c TransformationsWithout spelling out the Dirac components,

…where n = u, d, s, c, b, t indicates the “&avor.”Arranging the colors in a matrix format,

the quark wave-function phase becomes 3×3 matrix-valued,as does the unitary phase-transformation operator Uφ.

where Qa are 3×3 matricesHermitian, so Uφ would be unitary,traceless, so Uφ would be unimodular.

8

Yn(x) ! eigcjj

j

j

(x)/h̄ Yn(x), jj

j

j

(x) := j

a(x)Qa,

Diagonal phase-transformationpertains to electromagnetism…

Yn(x) = ê

a

Ya

n(x) = êrYrn(x) + êyYy

n(x) + êbYbn(x) =x) =

"Yr

n(x)Yy

n(x)Yb

n(x)

#,


The SU(3)c TransformationsGauge (local symmetry) transformations

Notice the multi-component-ness:

…which is usually suppressed in notation.In general:

where however the form of Qa depends on what it acts upon.9

Matrix-valued phases and local symmetry

[ih̄ /D � mc]Yn(x) = 0 ! [ih̄ /D0 � mc]Y0n(x) = 0,

0 �1

! �D

µ

! D0µ

:= Ujj

j

j

Dµ

U�1

jj

j

j

,

ig /h̄U

jj

j

j

:= eigcjj

j

j

/h̄,

/DYn ⌘ gg

g

g

µDµ

Yn, ( /DYn)a ⌘ gg

g

g

µDµ

a

b

Yb

n, ( /DYn)aA ⌘ (gµ)A

BDµ

a

b

YbBn ,

Dµ

:= 1 ∂

µ

+ igch̄ c Aa

µ

Qa,

a a 1 h̄ c


a,b,c = 1,…,8

Matrix representations of SU(3)

The SU(3)c TransformationsAs obtained in the “general” formalism:

…where

…and where

are the 3×3 matrices that act upon the (quark) color 3-vector.But, what about the Qa’s acting on the 8 Aμ

a’s or 8 φa’s?

10

h̄ c

A0aµ

Qa = Aaµ

Ujj

j

j

QaU�1

jj

j

j

+ h̄ cigc

Ujj

j

j

(∂µ

U�1

jj

j

j

) = Aaµ

Ujj

j

j

QaU�1

jj

j

j

� c(∂µ

j

a)Qa,

i.e. 0 �1 : a

A0µ

= Ujj

j

j

Aµ

U�1

jj

j

j

� c(∂µ

jj

j

j

), Aµ

:= Aaµ

Qa,

[Qa, Qb] = i fabc Qc.

d Aaµ

= �(Dµ

jj

j

j

)a := �c(∂µ

j

a) + igch̄ c Ab

µ

( eQb)ca

j

c =a a ac = �(∂

µ

j

a)� gch̄ c Ab

µ

fbca

j

c,

.

µ

�( eQb)c

a = i fbca,


nonlinear

The curvature tensor and the Bianchi identity

The SU(3)c-invariant Lagrangian

Notice the differences: D′μ = Uφ Dμ Uφ–1 implies

Also,

But, for the non-abelian (matrix-valued) case:

Recall however that

11

for electromagnetism,

Fµn

(A0) = Fµn

(A),

�∂

µ

(A0)an

� ∂

n

(A0)aµ

�6= (∂

µ

Aan

� ∂

n

Aaµ

),

� ��∂

µ

(A0)an

� ∂

n

(A0)aµ

�6= U

jj

j

j

(∂µ

Aan

� ∂

n

Aaµ

)U�1

jj

j

j

.

Note, however, that both in electrodynamics and in chromodynamics the derivatives: D0

µ

= Ujj

j

j

Dµ

U�1

jj

j

j

,

) A0µ

= Aµ

� (∂µ

j)

) (A0)aµ

= Aaµ

� (Dµ

j

a) = Aaµ

� (∂µ

j

a) + gch̄ c Ab

µ

fbca

j

c.




In electrodynamics:

It must be a commutator, so that the result would not be a differential operator, but an “ordinary” function.A commutator also computes the mismatch in… well,…commuting.In general: [D, D] = (torsion)·D + (curvature).So:

12

⇥D

µ

, Dn

⇤=

⇥∂

µ

+ iqh̄ c A

µ

, ∂

n

+ iqh̄ c A

n

⇤= + iq

h̄ c (∂µ

An

� ∂

n

Aµ

) = iqh̄ c F

µn

.

Fµn

:= h̄ cigc

⇥D

µ

, Dn

⇤= h̄ c

igc

⇥∂

µ

+ igch̄ c Ab

µ

Qb , ∂

n

+ igch̄ c Ac

n

Qc⇤,

ig

⇥ ⇤ ⇥ ⇤= (∂

µ

Aan

� ∂

n

Aaµ

)Qa + h̄ cigc

( igch̄ c )

2 Abµ

Acn

[Qb, Qc] = Faµn

Qa,

a a g a b c

�

Faµn

:= (∂µ

Aan

� ∂

n

Aaµ

)� gch̄ c f a

bc Abµ

Acn

,


Curvature and torsion

DigressionPicture the result of computing [D, D′] = DD′–D′D…in the total space of a #ber bundle:

13

#ber

base

base

D′

D

D D′

D′base

Dbase

Dbase

D′base

torsion

curvature

torsion

Generally, the rate-of-change operators (D) will fail to commute both in the#ber-space direction(= curvature: at the same “place” but of a different “value”) and in thebase-space direction(= torsion: not even at the same “place”).




Of course, we have that

Independently,

And, for all SU(n):

14

Fµn

! F0µn

:= ih̄ cgc

⇥D0

µ

, D0n

⇤= ih̄ c

gc

⇥U

jj

j

j

Dµ

U�1

jj

j

j

, Ujj

j

j

Dn

U�1

jj

j

j

⇤1 ⇤

= ih̄ cgc

Ujj

j

j

⇥D

µ

, Dn

⇤U�1

jj

j

j

,

c

⇥= U

jj

j

j

Fµn

U�1

jj

j

j

.

Dµ

(Fnr

) = [Dµ

, Fµn

] = h̄ cigc

⇥D

µ

, [Dn

, Dr

]⇤.⇥

A , [ B , C ]⇤+

⇥B , [C , A ]

⇤+

⇥C , [ A , B ]

⇤⌘ 0,

#

µnrsDµ

(Fnr

) = h̄ cigc

#

µnrs

⇥D

µ

, [Dn

, Dr

]⇤= 0,

Tr[Fµn

] = Fkµn

Tr[Qk] = 0.


Equations of motion


Since the matrix-valued (su(3) algebra-valued) curvaturetransforms by similarity transformation,…with respect to which the trace function is invariant,

…so one chooses:

15

Tr[Fµn

Fµn] ! Tr[F0µn

F0 µn] = Tr[Ujj

j

j

Fµn

U�1

jj

j

j

Ujj

j

j

FµnU�1

jj

j

j

] =

Tr

µn] = Tr[Fµn

FµnU�1

jj

j

j

Ujj

j

j

],

(4.22)= Tr[Fµn

Fµn]

LQCD = Ân

Tr

⇥

Yn(x) [ih̄ c/

D�mnc2]Yn(x)⇤

� 1

4

Tr

⇥

FµnFµn⇤,

h

� �

i

⇥ ⇤ ⇥ ⇤

= Ân

Ya n(x)h

iggggµ�h̄ cdab∂µ+igc Aa

µ( 1

2

la)a

b�

�mnc2dab

i

Ybn(x)

� 1

4

FaµnFµn

a .



Variation by Aaμ yields:

But, while

implies

the same is not true of Dμ Fa μν.Instead:

16

Equations of motion

(DµFµn = ∂µFµn) = geYA(gn)A

BYA,

∂

n

jn

e =4pe

0

c4p

∂

n

∂

µ

Fµn ⌘ 0, since Fµn

= �Fnµ

Dn ja n(q) = Dn Dµ Fa µn = � 1

2

[Dµ, Dn]Fa µn = � 1

2

f abcFb

µnFc µn = 0,

Dµ

Fa µn = gc Ân

YnaA(gn)A

B( 1

2

l

a)a

b

YbBn ,

ja µ

(q) := gc Ân

YnaA(gµ)A

B( 1

2

l

a)a

b

YbBn .

However, the left-hand side of Eq. (4.25) contains terms non-linear in


Color conservation and equation of continuity


!is does not lead to a conserved color:

…as the additional right-hand side term doesn’t vanish.However, use that

…so both quarks and gluons contribute to the color charge:

17

0 = Dµ ja µ(q) = ∂µ ja µ

(q) �gch̄ c f a

bc Abµ jc µ

(q) ,

) d

dt

⇣

Z

Vd

3~r ja 0

(q)

⌘

= �I

∂Vd

2~r ·~ja(q) +

gch̄ c f a

bc

⇣

Z

Vd

3~r Abµ jc µ

(q)

⌘

,

Dµ

Fa µn = ∂

µ

Fa µn � gch̄ c fbc

a Abµ

Fc µn

,

Dµ

Fa µn = ja n

(q) ) ∂

µ

Fa µn = Ja n

(c) , ) ∂

n

Ja n

(c) = 0,

Ja n

(c) := ja µ

(q) +gch̄ c fbc

a Abµ

Fc µn

,Z ZQa

(c) :=Z

d

3~r Ja 0

(c) = gc

Zd

3~r�Ân[Yn gg

g

g

µ

1

2

ll

l

l

a Yn] + 1

h̄ c fbca Ab

µ

Fc µn

�,




!is changes the analogues of Gauss-Ampère laws.Consider the ν = 0 case of the equation Dμ Fa μν = j(q)

aν:

…and de#ne:

!en,

andit is impossible to write analogues of Maxwell’s equationswith no reference to the gauge potentialsboth quarks and gluons serve as “sources” for the color force-#eldthe equations are nonlinear.

18

∂

µ

Fa µ0 � gch̄ c f a

bc Abµ

Fc µ0 = ja 0

(q) ,

~Ea := êiFa i0, r

a(q) := ja 0

(q) ,

~Aa := �ê

i Aai ,

~r·~Ea = r

a(q) �

gch̄ c f a

bc ~Ab·~Ec,




To sum up:

are the matrix-valued analogues ofthe Maxwell’s equationsthe conserved color-current & color-charge.

19

Dµ

Fµn = Jn

(q) and #

µnrsDµ

(Fnr

) = 0,

∂

µ

Fµn = Jn

(c),

RJn

(c) := Jn

(q) +igch̄ c [Aµ

, Fµn],

∂

n

Jn

(c) = 0,

d

dt

ZV

d

3~r J0

(c) = �I

∂Vd

2~s ·~J(c).

It should be clear that the application to an abelian (commutative) group where

Jn

(q) := gc

⇣Ân

YnaA(gµ)A

B( 1

2

l

a)a

b

YaAn

⌘Qa


Thanks!

Tristan HubschDepartment of Physics and Astronomy

Howard University, Washington DCPrirodno-Matematički Fakultet

Univerzitet u Novom Sadu

http://homepage.mac.com/thubsch/




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(Fundamental) Physics of Elementary Particlesphysics1.howard.edu/~thubsch/FPP1/Slides/1031.pdf · Department of Physics and Astronomy Howard University, Washington DC ... Fundamental